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# 7.4: Solving Multiplication and Division Equations

Difficulty Level: At Grade Created by: CK-12

## Introduction

Figuring Fares

While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass.

“How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign.

“We have $45.00 each allotted for the subway,” she said. “Well, a day pass is$9.00, but a month long pass is $20.00. I think that is a better deal,” he told her. “Let’s figure out how many days we can ride for$9.00 per day.”

Kara wrote the following equation on a piece of paper.

\begin{align*}9x=45\end{align*}

Solving equations is essential for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go.

You can figure this out too. Solving equations is what you will learn in this lesson. Take what you learn and then apply it back to this problem. At the end of the lesson, you will know which train option makes the most sense.

What You Will Learn

In this lesson you will learn the following skills:

• Solve single-variable multiplication equations.
• Solve single-variable division equations.
• Recognize and apply the multiplication and division properties of equality and the inverse relationship of multiplication and division.
• Model and solve real-world problems using multiplication and division equations.

Teaching Time

I. Solve Single-Variable Multiplication Equations

As you saw in the last lesson, we can have equations with addition and subtraction in them. We can also have equations with multiplication and division in them. These single-variable equations can be solved for the value of the variable. We want to solve the equation and figure out what the variable is equal to.

How does this work?

In the algebraic equation below, the variable \begin{align*}z\end{align*} represents only one possible number.

\begin{align*}z \times 2=8\end{align*}

What number does \begin{align*}z\end{align*} represent? We can find out by asking ourselves: what number, when multiplied by 2, equals 8?

Since \begin{align*}4 \times 2=8, z\end{align*} must be equal to 4. We solved this equation using mental math.

When we determine the value for a variable in an equation, we are solving that equation. If the equation involves a simple number fact, such as \begin{align*}4 \times 2=8\end{align*}, then we can solve the problem as we did above using mental math.

What happens when we have a more challenging equation to solve?

Some equations are more complex and mental math might not get the job done. To solve a more complex equation, such as \begin{align*}z \times 7=105\end{align*}, we might need to use a different strategy. Let's take a look at one strategy for solving an equation now.

One strategy for solving an equation is to use inverse operations to isolate the variable. Remember, isolating the variable means getting the variable by itself on one side of the equal (=) sign.

To solve an equation in which a variable is multiplied by a number, we can use the inverse operation of multiplication: division. We can divide both sides of the equation by that number to find the value of the variable.

We must divide both sides of the equation by that number because of the Division Property of Equality, which states:

if \begin{align*}a=b\end{align*} and \begin{align*}c \neq 0\end{align*}, then \begin{align*}\frac{a}{c}=\frac{b}{c}\end{align*}.

This is much simpler than it looks. If you divide one side of an equation by a nonzero number, \begin{align*}c\end{align*}, you must divide the other side of the equation by that same number, \begin{align*}c\end{align*}, to keep the values on both sides equal.

Example

\begin{align*}z \times 7=105\end{align*}.

In the equation, \begin{align*}z\end{align*} is multiplied by 7. So, we can divide both sides of the equation by 7 to solve for \begin{align*}z\end{align*}.

\begin{align*}z \times 7 &= 105\\ \frac{z \times 7}{7} &= \frac{105}{7}\\ z \times \frac{7}{7} &= 15\\ z \times 1 &= 15\\ z &= 15\end{align*}

It may help to separate out the factors like we did above.

Having divided both sides by 7, we can see that the value of \begin{align*}z\end{align*} is 15.

Example

\begin{align*}-8r=128\end{align*}

In the equation, -8 is multiplied by \begin{align*}r\end{align*}. So, we can divide both sides of the equation by -8 to solve for \begin{align*}r\end{align*}.

\begin{align*}-8r &= 128\\ \frac{-8r}{-8} &= \frac{128}{-8}\end{align*}

We need to use what we know about dividing integers to help us solve this problem.

\begin{align*}\frac{-8r}{-8} &= \frac{128}{-8}\\ 1r &= -16\\ r &= -16\end{align*}

The value of \begin{align*}r\end{align*} is -16 .

Remember the integer rules for dividing positive and negative numbers? Let’s review.

Now you can practice a few of these on your own.

7I. Lesson Exercises

Solve each problem for the missing variable.

1. \begin{align*}-4x=12\end{align*}
2. \begin{align*}8a=64\end{align*}
3. \begin{align*}9b=81\end{align*}

II. Solve Single-Variable Division Equations

Sometimes, you will see equations that have division in them. Remember, that we can use a fraction bar to show division.

To solve an equation in which a variable is divided by a number, we can use the inverse of division––multiplication. We can multiply both sides of the equation by that number to solve it.

We must multiply both sides of the equation by that number because of the Multiplication Property of Equality, which states:

if \begin{align*}a=b\end{align*}, then \begin{align*}a \times c=b \times c\end{align*}.

So, if you multiply one side of an equation by a number, \begin{align*}c\end{align*}, you must multiply the other side of the equation by that same number, \begin{align*}c\end{align*}, to keep the values on both sides equal.

Example

\begin{align*}k \div (-4)=12\end{align*}.

In the equation, \begin{align*}k\end{align*} is divided by -4. So, we can multiply both sides of the equation by -4 to solve for \begin{align*}k\end{align*}. You will need to use what you know about multiplying integers to help you solve this problem. It may help to rewrite \begin{align*}k \div (-4)\end{align*} as \begin{align*}\frac{k}{-4}\end{align*}.

\begin{align*}k \div (-4) &= 12\\ \frac{k}{-4} &= 12\\ \frac{k}{-4} \times (-4) &= 12 \times (-4)\\ \frac{k}{-4} \times \frac{-4}{1} &= -48\\ \frac{k}{\cancel{-4}} \times \frac{\cancel{-4}}{1} &= -48\\ \frac{k}{1} &= -48\\ k &= -48\end{align*}

The –4's will cancel each other out when they are divided. Then we multiply.

The value of \begin{align*}k\end{align*} is –48.

Remember the rules for multiplying integers will apply when working with these equations!! Think back and use them as you work.

Example

\begin{align*}\frac{n}{1.5}=10\end{align*}

In the equation, \begin{align*}n\end{align*} is divided by 1.5. So, we can multiply both sides of the equation by 1.5 to solve for \begin{align*}n\end{align*}.

\begin{align*}\frac{n}{1.5} &= 10\\ \frac{n}{1.5} \times 1.5 &= 10 \times 1.5\\ \frac{n}{1.5} \times \frac{1.5}{1} &= 15\\ \frac{n}{\cancel{1.5}} \times \frac{\cancel{1.5}}{1} &= 15\\ \frac{n}{1} &= 15\\ n &= 15\end{align*}

The value of \begin{align*}n\end{align*} is 15.

Now it is time for you to practice solving a few of these equations.

7J. Lesson Exercises

Solve each equation for the missing variable.

1. \begin{align*}\frac{x}{-2}=5\end{align*}
2. \begin{align*}\frac{y}{5}=6\end{align*}
3. \begin{align*}\frac{b}{-4}=-3\end{align*}

Take a few minutes to check your work with a peer.

III. Using and Understanding Properties

In the last two sections, we have been using properties to help us in solving equations.

Remember that a property is a rule for performing operations in mathematics that makes our lives simpler. These properties are no exception. If you can look at an equation, see that it is a multiplication equation and know that division is the way to solve it, then the property of the Division Property of Equality is being utilized.

In the same way if you use the Multiplication Property of Equality to solve a division problem, then the property has been useful. Always remember to think about the inverse operations and associate them with the different properties. This will help you to keep it all straight and not get mixed up.

The Multiplication and Division Properties of Equality in combination with inverse operations are the best ways to solve equations correctly.

IV. Model and Solve Real-World Problems Using Multiplication and Division Equations

Equations are very useful when working with real-world problems. In fact, an equation is often the best way to sift through the information in a problem, select the correct information and work to solve an equation to find an answer.

Let’s apply what we have learned to some real-life dilemmas.

Example

Three friends evenly split the total cost of the bill for their lunch. The amount each friend paid for his share was 4.25. a. Write an equation to represent \begin{align*}c\end{align*}, the total cost, in dollars, of the bill for lunch. b. Determine the total cost of the bill. Consider part \begin{align*}a\end{align*} first. Use a number, an operation sign, a variable, or an equal sign to represent each part of the problem. Because the total cost was evenly split by 3 friends, write a division equation to represent the problem. \begin{align*}& \underline{Three \ friends} \ \underline{evenly \ split} \ the \ \underline{total \ cost} \ldots amount \ each \ friend \ paid \ \underline{was} \ \underline{\4.25} \ldots\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ \Box \qquad \qquad \qquad \downarrow \qquad \qquad \quad \Box \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \downarrow \qquad \downarrow\\ & \qquad \ c \qquad \qquad \qquad \ \div \qquad \qquad \quad 3 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ = \quad 4.25\end{align*} This equation, \begin{align*}c \div 3=4.25\end{align*}, represents \begin{align*}c\end{align*}, the total cost of the lunch bill. Next, consider part \begin{align*}b\end{align*}. Solve the equation to find the total cost, in dollars, of the lunch bill. \begin{align*}c \div 3 &= 4.25\\ \frac{c}{3} &= 4.25\\ \frac{c}{3} \times 3 &= 4.25 \times 3\\ \frac{c}{\cancel{3}} \times \frac{\cancel{3}}{1} &= 12.75\\ c &= 12.75\end{align*} The total cost of the lunch bill was12.75.

Example

Sarvenaz earns $8 for each hour she works. She earned a total of$168 last week.

a. Write an equation to represent \begin{align*}h\end{align*}, the number of hours she worked last week.

b. Determine how many hours Sarvenaz worked last week.

Consider part \begin{align*}a\end{align*} first.

Use a number, an operation sign, a variable, or an equal sign to represent each part of that problem. She earns $8 for each hour she works, so you could multiply the number of hours she worked by$8 to find the total amount she earned. Write a multiplication equation.

\begin{align*}& \underline{\8 \ for \ each \ hour}\ldots She \ \underline{earned} \ a \ \underline{total \ of \ \168} \ last \ week.\\ & \qquad \downarrow \qquad \qquad \qquad \qquad \qquad \downarrow \qquad \qquad \ \downarrow\\ & \qquad 8h \qquad \qquad \qquad \qquad \quad \ = \qquad \quad \ 168\end{align*}

So, this equation, \begin{align*}8h=168\end{align*}, represents \begin{align*}h\end{align*}, the number of hours she worked last week.

Next, consider part \begin{align*}b\end{align*}.

Solve the equation to find \begin{align*}h\end{align*}, the number of hours she worked last week.

\begin{align*}8h &= 168\\ \frac{8h}{8} &= \frac{168}{8}\\ 1h &= 21\\ h &= 21\end{align*}

Sarvenaz worked 21 hours last week.

Now let’s go back and apply what we have learned to the introduction problem.

## Real Life Example Completed

Figuring Fares

Here is the original problem once again. Reread it and underline any important information.

While visiting their grandparents, Kara has been handling the money for their rides on the subway. For the past two days, Marc has been watching her calculating and handing out money. Then when they were coming back from the mall, Marc noticed something on a sign about a day pass and a month long pass.

“How much money do we have set aside for the subway?” Marc asked Kara the day after he saw the sign.

We have $45.00 each allotted for the subway,” she said. “Well, a day pass is$9.00, but a month long pass is $20.00. I think that is a better deal,” he told her. “Let’s figure out how many days we can ride for$9.00 per day.”

Kara wrote the following equation on a piece of paper.

\begin{align*}\underline{9x=45}\end{align*}

Solving equations is essential for Kara and Marc at this point in time. Then they will know whether day passes or a month long pass is the way to go.

First, let’s solve the equation and figure out how many day passes Kara and Marc can purchase day passes given their budget.

\begin{align*}9x &= 45\\ \frac{9x}{9} &= \frac{45}{9}\\ x &= 5\end{align*}

According to this, they can purchase day passes for only five days.

A month long pass is 20.00 each. This is a much better deal for them. \begin{align*}20(2) = \40.00\end{align*} The total subway fares will cost them40.00 together instead of \$45.00 each.

Kara and Marc purchase month long T passes the very next morning.

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Isolate the Variable
get the variable alone on one side of the equals sign.
Inverse Operation
Opposite operation
Division Property of Equality
states that if both sides of an equation are divided by the same number, they will stay equal.
Multiplication Property of Equality
states that if both sides of an equation are multiplied by the same number, they will stay equal.

## Technology Integration

Other Videos:

1. http://www.mathplayground.com/mv_solving_single_step_equations.html – This is a Brightstorm video about solving single step equations.

## Time to Practice

Directions: Solve each single-variable multiplication equation for the missing value.

1. \begin{align*}4x=16\end{align*}

2. \begin{align*}6x=72\end{align*}

3. \begin{align*}-6x=72\end{align*}

4. \begin{align*}-3y=24\end{align*}

5. \begin{align*}-3y=-24\end{align*}

6. \begin{align*}-5x=-45\end{align*}

7. \begin{align*}-1.4x=2.8\end{align*}

8. \begin{align*}3.5a=7\end{align*}

9. \begin{align*}7a=-49\end{align*}

10. \begin{align*}14b=-42\end{align*}

Directions: Solve each single-variable division equation for the missing value.

11. \begin{align*}\frac{x}{5}=2\end{align*}

12. \begin{align*}\frac{y}{7}=3\end{align*}

13. \begin{align*}\frac{b}{9}=-4\end{align*}

14. \begin{align*}\frac{b}{8}=-10\end{align*}

15. \begin{align*}\frac{x}{3}=-3\end{align*}

16. \begin{align*}\frac{x}{5}=-8\end{align*}

17. \begin{align*}\frac{x}{1.3}=3\end{align*}

18. \begin{align*}\frac{x}{2.4}=4\end{align*}

19. \begin{align*}\frac{x}{6}=1.2\end{align*}

20. \begin{align*}\frac{y}{1.5}=3\end{align*}

Directions: Solve each problem.

21. Mrs. Kowalski divided the students in the seventh grade into 7 equal-sized groups. There were 25 students in each group.

a. Write an equation to represent \begin{align*}s\end{align*}, the total number of students in the seventh grade.

b. Determine the total number of students in the seventh grade.

22. Evan needs to buy a total of 180 hot dogs for a school event. Hot dogs are sold in packages of 12 each.

a. Write an equation to represent \begin{align*}p\end{align*}, the number of packages Evan must buy.

b. Determine the number of hot dog packages Evan must buy.

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